How many terms are there in the AP 6, 10, 14, 18,...,174?
Question: How many terms are there in the AP 6, 10, 14, 18,...,174? Solution: In the given AP,a= 6 andd= (10 - 6) = 4Suppose that there arenterms in the given AP.Then Tn= 174⇒a+ (n- 1)d = 174⇒ 6 + (n- 1) ⨯ 4 = 174⇒ 2 + 4n= 174⇒ 4n= 172⇒n= 43Hence, there are 43 terms in the given AP....
Read More →If f(x) and g(x) are two polynomials such
Question: If $f(x)$ and $g(x)$ are two polynomials such that the polynomial $\mathrm{P}(\mathrm{x})=f\left(\mathrm{x}^{3}\right)+\mathrm{xg}\left(\mathrm{x}^{3}\right)$ is divisible by $x^{2}+x+1$, then $P(1)$ is equal to________. Solution: $P(x)=f\left(x^{3}\right)+x g\left(x^{3}\right)$ $P(1)=f(1)+g(1) \ldots(1)$ Now $\mathrm{P}(\mathrm{x})$ is divisible by $\mathrm{x}^{2}+\mathrm{x}+1$ $\Rightarrow \mathrm{P}(\mathrm{x})=\mathrm{Q}(\mathrm{x})\left(\mathrm{x}^{2}+\mathrm{x}+1\right)$ $\mathrm...
Read More →If the nth term of a progression is (4n − 10), show that it is an AP. Find its
Question: If thenth term of a progression is (4n 10), show that it is an AP. Find its(i) first term, (ii) common difference, and (iii) 16th term. Solution: Tn= (4n- 10) [Given]T1= (4 ⨯ 1 - 10) = -6T2= (4 ⨯ 2 - 10) = -2T3= (4 ⨯ 3 - 10) = 2T4= (4 ⨯ 4 - 10) = 6 Clearly, [ -2 - (-6)] = [2 - (-2)] = [6 - 2] = 4 (Constant) So, the terms -6, -2, 2, 6,... forms an AP.Thus, we have;(i) First term = -6 (ii) Common difference = 4(iii) T16=a+ (n-1)d =a+ 15d= - 6 + 15 ⨯ 4 = 54...
Read More →Find the nth term of each of the following APs:
Question: Find thenth term of each of the following APs:(i) 5, 11, 17, 23, ....(ii) 16, 9, 2, 5, .... Solution: (i) $5,11,17,23, \ldots$ $a_{n}=a+(n-1) d$ $a=5$ $d=11-5=6$ $a_{n}=5+(n-1) 6$ $a_{n}=6 n-1$ (ii) $16,9,2,-5$ $a_{n}=a+(n-1) d$ $a=16$ $d=9-16=-7$ $a_{n}=16+(n-1)(-7)$ $a_{n}=16-7 n+7$ $a_{n}=23-7 n$...
Read More →Find the value of p for which the numbers 2p – 1, 3p + 1, 11 are in AP.
Question: Find the value ofpfor which the numbers 2p 1, 3p+ 1, 11 are in AP. Hence, find the numbers. Solution: $2 p-1,3 p+1,11$ are in $\mathrm{AP}$ A.P has common difference $d=a_{2}-a_{1}$ Here, $a_{1}=2 p-1, a_{2}=3 p+1$ and $a_{3}=11$ $a_{2}-a_{1}=a_{3}-a_{2}$ $3 p+1-(2 p-1)=11-3 p-1$ $p+2=10-3 p$ $\Rightarrow p=2$ And the numbers are: $2(2)-1,3(2)+1,11$ $\Rightarrow 3,7,11$ are the numbers...
Read More →The value of
Question: The value of $3+\frac{1}{4+\frac{1}{3+\frac{1}{4+\frac{1}{3+\ldots \infty}}}}$ is equal to (1) $1.5+\sqrt{3}$(2) $2+\sqrt{3}$(3) $3+2 \sqrt{3}$(4) $4+\sqrt{3}$Correct Option: 1 Solution: Let $x=3+\frac{1}{4+\frac{1}{3+\frac{1}{4+\frac{1}{3+\ldots \infty}}}}$ So, $x=3+\frac{1}{4+\frac{1}{x}}=3+\frac{1}{\frac{4 x+1}{x}}$ $\Rightarrow(x-3)=\frac{x}{(4 x+1)}$ $\Rightarrow(4 x+1)(x-3)=x$ $\Rightarrow 4 x^{2}-12 x+x-3=x$ $\Rightarrow 4 x^{2}-12 x-3=0$ $\mathrm{x}=\frac{12 \pm \sqrt{(12)^{2}+...
Read More →The major product of the following reaction is:
Question: The major product of the following reaction is: Correct Option: , 3 Solution:...
Read More →Proton with kinetic energy
Question: Proton with kinetic energy of $1 \mathrm{MeV}$ moves from south to north. It gets an acceleration of $10^{12} \mathrm{~m} / \mathrm{s}^{2}$ by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is $1.6 \times 10^{-27} \mathrm{~kg}$ )(1) $0.71 \mathrm{mT}$(2) $7.1 \mathrm{mT}$(3) $0.071 \mathrm{mT}$(4) $71 \mathrm{mT}$Correct Option: 1 Solution: As we know, magnetic force $F=q v B=m a$ $\therefore \vec{a}=\left(\frac{q v B}{m}\right)$ perpendicul...
Read More →Find the 37th term of the AP
Question: (i) Find the 37 th term of the AP $6,7 \frac{3}{4}, 9 \frac{1}{2}, 11 \frac{1}{4}, \ldots$ (ii) Find the 25 th term of the AP $5,4 \frac{1}{2}, 4,3 \frac{1}{2}, 3, \ldots$ Solution: (i) The given AP is $6,7 \frac{3}{4}, 9 \frac{1}{2}, 11 \frac{1}{4}, \ldots$ First term, $a=6$ and common difference, $d=7 \frac{3}{4}-6 \Rightarrow \frac{31}{4}-6 \Rightarrow \frac{31-24}{4}=\frac{7}{4}$ Now, $T_{37}=a+(37-1) d=a+36 d$ $=6+36 \times \frac{7}{4}=6+63=69$ $\therefore 37^{t h}$ term $=69$ (ii...
Read More →The value of
Question: The value of $4+\frac{1}{5+\frac{1}{4+\frac{1}{5 \cdot \frac{1}{4+\ldots \infty}}}}$ (1) $2+\frac{2}{5} \sqrt{30}$(2) $2+\frac{4}{\sqrt{5}} \sqrt{30}$(3) $4+\frac{4}{\sqrt{5}} \sqrt{30}$(4) $5+\frac{2}{5} \sqrt{30}$Correct Option: 1 Solution: $y=4+\frac{1}{\left(5+\frac{1}{y}\right)}$ $y-4=\frac{y}{(5 y+1)}$ $5 y^{2}-20 y-4=0$ $y=\frac{20+\sqrt{480}}{10}$ $y=\frac{20-\sqrt{480}}{10} \rightarrow$ rejected $\mathrm{y}=2+\sqrt{\frac{480}{100}}$ Correct with Option (A)...
Read More →The major product of the following reaction is:
Question: The major product of the following reaction is: Correct Option: , 3 Solution:...
Read More →A loop ABCDEFA of straight edges has six corner points
Question: A loop ABCDEFA of straight edges has six corner points $A(0,0,0), B\{5,0,0), C(5,5,0), D(0,5,0), E(0,5,5)$ and $F(0,0,5)$. The magnetic field in this region is $\vec{B}=(3 \hat{i}+4 \hat{k}) \mathrm{T}$. The quantity of flux through the loop $A B C D E F A$ (in $\mathrm{Wb}$ ) is_______ Solution: Flux through the loop ABCDEFA, $\phi=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}=(3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}}) \cdot(25 \hat{\mathrm{i}}+25 \hat{\mathrm{k}})$ $\Rig...
Read More →Prove the following
Question: Let $\mathrm{P}(\mathrm{x})=\mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}$ be a quadratic polynomial with real coefficients such that $\int_{0}^{1} \mathrm{P}(\mathrm{x}) \mathrm{dx}=1$ and $\mathrm{P}(\mathrm{x})$ leaves remainder 5 when it is divided by $(x-2)$. Then the value of $9(b+c)$ is equal to:(1) 9(2) 15(3) 7(4) 11Correct Option: , 3 Solution: $\int_{0}^{1}\left(\mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}\right) \mathrm{dx}=1$ $\frac{1}{3}+\frac{b}{2}+c=1 \quad \Rightarrow \quad \frac{b}{2...
Read More →Find:
Question: Find: (i) the 20 th term of the AP $9,13,17,21, \ldots .$ (ii) the 35 th term of the AP $20,17,14,11$, (iii) the 18 th term of the AP $\sqrt{2}, \sqrt{18}, \sqrt{50}, \sqrt{98}, \ldots$. (iv) the 9 th term of the AP $\frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \frac{9}{4}, \ldots .$ (v) the 15 th term of the AP $-40,-15,10,35, \ldots .$ Solution: (i) The given AP is 9, 13, 17, 21, ... .First term,a= 9Common difference,d= 13 9 = 4nthterm of the AP,an=a+ (n 1)d= 9 +(n 1) 420th term of the AP,...
Read More →Consider a circular coil of wire carrying constant current I
Question: Consider a circular coil of wire carrying constant current $I$, forming a magnetic dipole. The magnetic flux through an infinite plane that contains the circular coil and excluding the circular coil area is given by $\phi_{i}$.The magnetic flux through the area of the circular coil area is given by $\phi_{0}$. Which of the following option is correct?(1) $\phi_{i}=\phi_{0}$(2) $\phi_{i}\phi_{0}$(3) $\phi_{i}\phi_{0}$(4) $\phi_{i}=-\phi_{0}$Correct Option: , 4 Solution: (4) As magnetic ...
Read More →The number of elements in the set
Question: The number of elements in the set $\{x \in \mathbb{R}:(|x|-3)|x+4|=6\}$ is equal to(1) 3(2) 2(3) 4(4) 1Correct Option: , 2 Solution: $x \neq-4$ $(|x|-3)(|x+4|)=6$ $\Rightarrow \quad|x|-3=\frac{6}{|x+4|}$ No. of solutions $=2$...
Read More →The major product of the following reaction is:
Question: The major product of the following reaction is: Correct Option: 1 Solution:...
Read More →A square loop of side $2 a$ and carrying current I is kept is x z plane with its centre at origin.
Question: A square loop of side $2 a$ and carrying current $\mathrm{I}$ is kept is $x z$ plane with its centre at origin. A long wire carrying the same current $\mathrm{I}$ is placed parallel to $z$-axis and passing through point $(0, b, 0),(ba)$. The magnitude of torque on the loop about z-axis will be :(1) $\frac{2 \mu_{0} \mathrm{I}^{2} a^{2}}{\pi b}$(2) $\frac{2 \mu_{0} I^{2} a^{2} b}{\pi\left(a^{2}+b^{2}\right)}$(3) $\frac{\mu_{0} I^{2} a^{2} b}{2 \pi\left(a^{2}+b^{2}\right)}$(4) $\frac{\mu...
Read More →Show that each of the progressions given below is an AP. Find the first term, common difference and next term of each.
Question: Show that each of the progressions given below is an AP. Find the first term, common difference and next term of each. (i) $9,15,21,27, \ldots$ (ii) $11,6,1,-4$, (iii) $-1, \frac{-5}{6}, \frac{-2}{3}, \frac{-1}{2}, \ldots$ (iv) $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots$ (v) $\sqrt{20}, \sqrt{45}, \sqrt{80}, \sqrt{125}, \ldots$ Solution: (i) The given progression 9, 15, 21, 27, ... .Clearly, 15 9 = 21 15 = 27 21 = 6 (Constant)Thus, each term differs from its preceding term by 6....
Read More →The decreasing order of reactivity towards dehydrohalogenation
Question: The decreasing order of reactivity towards dehydrohalogenation $\left(E_{1}\right)$ reaction of the following compounds is: $\mathrm{D}\mathrm{B}\mathrm{C}\mathrm{A}$$\mathrm{B}\mathrm{D}\mathrm{A}\mathrm{C}$$\mathrm{B}\mathrm{D}\mathrm{C}\mathrm{A}$$BADC$Correct Option: 1 Solution: $\mathrm{E}_{1}$ reaction proceeds via carbocation formation, therefore greater the stability of carbocation, faster will be the $\mathrm{E}_{1}$ reaction. Thus correct decreasing order of the given halides...
Read More →Prove the following
Question: Let $A(1,0), B(6,2)$ and $\mathrm{C}\left(\frac{3}{2}, 6\right)$ be the vertices of a triangle $A B C$. If $P$ is a point inside the triangle $A B C$ such that the triangles $A P C, A P B$ and $B P C$ have equal areas, then the length of the line segment $P Q$, where $Q$ is the point $\left(-\frac{7}{6},-\frac{1}{3}\right)$, is_________. Solution: P will be centroid of $\triangle A B C$ $P\left(\frac{17}{6}, \frac{8}{3}\right) \Rightarrow P Q=\sqrt{(4)^{2}+(3)^{2}}=5$...
Read More →For the following reactions
Question: For the following reactions $\mu_{\mathrm{B}}\mu_{\mathrm{A}}$ and $k_{e}(\mathrm{~A})k_{e}(\mathrm{~B})$$\mu_{\mathrm{A}}\mu_{\mathrm{B}}$ and $k_{e}(\mathrm{~B})k_{e}(\mathrm{~A})$$\mu_{\mathrm{B}}\mu_{\mathrm{A}}$ and $k_{e}(\mathrm{~B})k_{e}(\mathrm{~A})$$\mu_{\mathrm{A}}\mu_{\mathrm{B}}$ and $k_{e}(\mathrm{~A})k_{e}(\mathrm{~B})$Correct Option: , 2 Solution: Among the given bases (A) and (B), $t$-butoxide being bulky base favours elimination reaction and ethoxide favours substitut...
Read More →A charged particle going around in a circle can be considered to be a current loop.
Question: A charged particle going around in a circle can be considered to be a current loop. A particle of mass $m$ carrying charge $q$ is moving in a plane wit speed $v$ under the influence of magnetic field $\vec{B}$. The magnetic moment of this moving particle :(1) $\frac{m v^{2} \vec{B}}{2 B^{2}}$(2) $-\frac{m v^{2} \vec{B}}{2 \pi B^{2}}$(3) $-\frac{m v^{2} \vec{B}}{B^{2}}$(4) $-\frac{m v^{2} \vec{B}}{2 B^{2}}$Correct Option: , 4 Solution: Length of the circular path, $l=2 \pi r$ Current, $...
Read More →Prove the following
Question: In $\Delta \mathrm{ABC}$, the lengths of sides $\mathrm{AC}$ and $\mathrm{AB}$ are $12 \mathrm{~cm}$ and $5 \mathrm{~cm}$, respectively. If the area of $\triangle \mathrm{ABC}$ is $30 \mathrm{~cm}^{2}$ and $\mathrm{R}$ and $\mathrm{r}$ are respectively the radii of circumcircle and incircle of $\triangle \mathrm{ABC}$ then the value of $2 R+r$ (in $c m$ ) is equal to____________. Solution: $\Delta=\frac{1}{2} \cdot 5 \cdot 12 \cdot \sin \mathrm{A}=30$ $\sin \mathrm{A}=1$ $\mathrm{A}=90...
Read More →Short-Answer Questions
Question: Short-Answer Questions Solve: $x^{2}-4 a x+4 a^{2}-b^{2}=0$ Solution: $x^{2}-4 a x+4 a^{2}-b^{2}=0$ $\Rightarrow x^{2}-4 a x+(2 a+b)(2 a-b)=0$ $\Rightarrow x^{2}-[(2 a+b)+(2 a-b)] x+(2 a+b)(2 a-b)=0$ $\Rightarrow x^{2}-(2 a+b) x-(2 a-b) x+(2 a+b)(2 a-b)=0$ $\Rightarrow x[x-(2 a+b)]-(2 a-b)[x-(2 a+b)]=0$ $\Rightarrow[x-(2 a+b)][x-(2 a-b)]=0$ $\Rightarrow x-(2 a+b)=0$ or $x-(2 a-b)=0$ $\Rightarrow x=(2 a+b)$ or $x=(2 a-b)$ Hence, (2a+b) and (2ab) are the roots of the given equation....
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